Question

evaluate.
integral [ ( 4x^3 ) / sqr(x^2 + 4) ] dx

Answer 1

\[ \int\frac{4x^3}{\sqrt{x^2+4}}dx\] Try this:
\[ \int\frac{2x(2x^2)}{\sqrt{x^2+4}}dx\]
Let \(x^2+4 = u \implies x^2=u-4\)
\(du = 2x\, dx\)

So by substitution:
\[ \int \frac{2(u-4)}{\sqrt{u}}\,du\]

Answer 2

sorry, i'm trying to use trig sub

Answer 3

If you could do trig sub, but @kirbykirby 's way is actually tons easier.
You want to base your trig sub off of that thing in the square root, the x^2+4 thing.

Answer 4

This trick is easier than trig sub I find :) , you can just separate the integral above in each separate integral should be easy to evaluate

Answer 5

i agree, but the instructions say use trig sub :(

Answer 6

It helps me to recall some of the trig identities (all of them is just a rewrite of the some identity, the Pythagorean identitity:
1-sin^2(x)=cos^2(x)
1+tan^2(x)=sec^2(x)
sec^2(x)-1=tan^2(x)

Answer 7

Do you see what trig identity will be useful here?

Answer 8

\[x^2+4=4(\frac{x^2}{4}+1)=4 ( (\frac{x}{2})^2+1)\]
And it might help if I factor out that 4
Let me rewrite that thing in the ( ) one more time:
\[1+(\frac{x}{2})^2 \]
compare this to all of those left sides of the identities I wrote above
1-sin^2(x)
1+tan^2(x)
sec^2(x)-1


which one looks the most similar

Answer 9

x = atan(theta)
dx = asec^2(theta)d(theta)

Answer 10

i have to go, thank you for your help

Answer 11

Since 1+tan^2(theta) would be the same as 1+(x/2)^2 if tan(theta)=x/2